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$I = \,\,\int\limits_{ - 1}^0 {\ln \left( {x + \sqrt {{x^2} + 1} }
\right)dx} + \,\,\int\limits_0^1 {\ln \left( {x + \sqrt {{x^2} + 1} }
\right)dx} $
$I = K + L$
Đặt $x = - t\,\,\,\,\,\,\, \Rightarrow \,\,\,\,dx\,\, = \,\, - dt$
$\begin{array}{l}
x = - 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,t\,\, = \,\,1\\
x = - t\,\,\,\,\,\,\, \Rightarrow \,\,\,\,t\,\, = \,\,0
\end{array}$
Suy ra :
$\begin{array}{l}
K
= \int\limits_{ - 1}^0 {\ln \left( {x + \sqrt {{x^2} + 1} } \right)dx =
} \int\limits_0^1 {\ln \left( { - t + \sqrt {{t^2} + 1} } \right)dt} \\
\,\,\,\,\,
= \,\,\,\int\limits_0^1 {\ln \frac{1}{{t + \sqrt {{t^2} + 1} }}} dt =
\,\, - \int\limits_0^1 {\ln \left( {x + \sqrt {{x^2} + 1} } \right)dx =
- L}
\end{array}$
Vậy $I = K + L = - L + L = 0$ ĐS : $I = 0$
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Trả lời 13-07-12 08:11 AM
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